C++ Trivia :

Some trivia for you ( while I wait for a compile of 22 projects to take place… ) why does the code below not compile ?

The Problem

class AnotherClass
{};
class SomeClass
{ 
public:
  SomeClass( AnotherClass anotherClass ){};
  SomeClass( int someint ){};
  void TheMethod(){};
};
int main(int argc, char* argv[] )
{
  AnotherClass anotherClass;
  SomeClass willCompile( anotherClass );
  willCompile.TheMethod();

  SomeClass wontCompile( AnotherClass() );
  wontCompile.TheMethod();//error C2228: left of '.TheMethod' must have class/struct/union 

  return;
}

Confused.. good then I am not the only one…Lets take a quick detour to something that appears to be totally unrelated.

Quick Detour

Now given that one of the design decision of C++ was that it would be a superset of C, it has to allow for you the ability to define function pointers. So for a simple method you can define a prototype like this in your code for a method that takes 1 param ( typically you would do this as part of a typedef )

int Function(int a); 

As a short cut you can define your prototype without any param names ( since they are meaningless at this level ). 

int Function(int); 

In the same way you can also define the prototype for a method that takes a function pointer as a param

void TakesMethod( void Function( int  a ) );

Taking advantage of the “shortcut” you can equally write 

void TakesMethod2( void ( int ) );

The issue

Right so after a quick C/C++ lesson.. can you see the issue… ? Let make it a bit more obvious.. lets define a method that take a function pointer to a method that takes 0 params

void      TakesMethodNoParams(  void          () );
SomeClass wontCompile        (  AnotherClass  () );

Yup we just confused the compiler… it thinks we are defining a prototype for a method takes a function pointer as its single param, which is a method that takes no params, and returns a type of Anotherclass